Let $X_n$ be independent Poisson random variables with $E[X_i] = \mu_i$, and let $Y_n = X_1+…+X_n$. I want to show that if $\sum_n \mu_n = \infty $ then $Y_n/E[Y_n] \rightarrow 1$ almost surly.
What I do know is that if $X_1,…$ are independent, and $E\left[X^4\right] < \infty$, then $Y_n/n \rightarrow E[X]$ a.s.* So What I tried is to center the random variables: let $C_n := X_n-\mu_n$. The $C$'s are independent, $E[C_n^4]< \infty, E[C_n] = 0$, which means:
$\frac{C_1 +…+C_n}{n} \rightarrow 0$ a.s., or $\frac{\sum X_i – \sum \mu_i}{n} \rightarrow 0$ a.s. which, I think, completes the proof.
However I am not sure why the requirement $\sum_n \mu_n = \infty $ is necessary, so I suspect this proof is incorrect. Is it?
Edit: This argument is incorrect since the theorem requires $E[C^4]$ to be uniformly bounded, not just bounded. Any ideas?
Best Answer
If, in your argument, $M=\sum_n \mu_n<\infty$ then your $Y_n$ converge in distribution to a Poisson rv. with expectation $M$, and your ratio $Y_n/EY_n$ has a non trivial limit distribution, violating a.s. convergence to a constant.
And, the theorem you quote needs a uniform 4th moment bound $EX_k^4\le K$, which need not hold given your hypotheses.
A better way to prove your result might be to write your $Y_n$ as $Y_n=Y(M_n)$, where $M_n=\sum_{t\le n} \mu_t$ and where $Y(t)$ is a Poisson process, as described in the wikipedia article. You might be able to show $Y(t)/t\to 1$ almost surely.
Alternatively, you could use your argument with the added hypothesis that $\mu_n\le1$ for all $n$ and then use that to prove the general result by representing each $\mu_n$ as a sum of quantities that are individually bounded by $1$, and your corresponding $Y_n$ as sums of independent Poissons whose expectations are bounded and add up to the right thing.